I already know the answer to be ke[tex]^{t^{2}/2}[/tex], but can't figure out how it got here. I'm rusty with my integrals and am just really starting diff eqs, but I think I'm just supposed to take the integral of ty (which is (t^2y)/2 i believe) and add a + c to the end, c being some constant, so I'm confused...

Staff: Mentor

The basic goal is to get y and dy on one side, and t and dt on the other, which is called separation. Your example differential equation is not separable, if I understand what you wrote. Is the first term in the denominator t^(2y) or is it t^2 * y? If it's the first, your example isn't separable.

These are called SEPARABLE DEs. Separable because we can separate them into f(y) dy = g(t) dt. And then integrate both sides. This is also why DE requires you to do well in Calculus classes so you don't have to think about how to integrate :) This is also why separable DEs are introduced in Calc 2(I guess), to give you a flavor of what to come.

In the case that you're uncertain how to integrate t dt/(t^2 + 1) use the substitution u = t^2 + 1. However, you really should learn how to do these steps and properly apply elementary algebra - especially is you're working with DEs.